Question

A solid cylinder has a mass of 100 kg and radius 0.225m. The cylinder is attached to a frictionless horizontal axle. A long (light weight) cable is wrapped around the cylinder. Attached to the end of the cable is a 1.50 kg mass. The system is initially stationary. The hanging mass is then released. The mass pulls on the cable as it falls and this causes the cylinder to rotate.

a) What is the velocity of the hanging mass after it has fallen 3.0 m?

b) what is the angular velocity (in rad/s) of the cylinder when the mass has fallen 3.0m?

c) What is the acceleration of the hanging mass and what is the angular acceleration of the rotating cylinder when the hanging mass has fallen 3.0m

d) find the tension in the cable.

Answer #1

a)

M = mass of cylinder = 100 kg

R = radius of cylinder = 0.225 m

w = angular speed of rotation

v = speed of hanging mass

h = height dropped = 3 m

m = mass of hanging mass = 1.50 kg

using conservation of energy

Potential energy = kinetic enery of hanging mass + rotational KE of cylinder

mgh = (0.5) m v^{2} + (0.5) I w^{2}

mgh = (0.5) m v^{2} + (0.5) (0.5) MR^{2}
(v/R)^{2}

mgh = (0.5) m v^{2} + (0.25) Mv^{2}

v = sqrt(mgh/((0.5)m + (0.25)M))

v = sqrt(1.50 x 9.8 x 3/((0.5) (1.5) + (0.25) (100)))

v = 1.31 m/s

b)

w = v/R = 1.31 /0.225 = 5.82 rad/s

c)

for the hanging mass :

Y = displacement = 3 m

a = acceleration

V_{i} = initial velocity = 0 m/s

V_{f} = final velocity = 1.31 m/s

using the equation

Vf^{2} = Vi^{2} + 2 a Y

1.31^{2} = 0^{2} + 2 a(3)

a = 0.286 m/s^{2}

= angular
acceleration = a/R = 0.286/0.225 = 1.27

d)

tension in the cable is given as

mg - T = ma

T = mg - ma

T = m(g - a) = 1.50 (9.8 - 0.286) = 14.3 N

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